# Is $\mathbb{R}^4$ minus a line simply connected?

Is the set $$\mathbb{R}^4\setminus \{(0,0,0,w) | w\in \mathbb{R} \}$$ simply connected? I started trying to grasp the notion of simple connectedness in higher dimensions and realized I could not even a figure out such a basic question.

My intuition says it is not simply connected but maybe you can twist things around in 4 dimensions in ways I can not visualize.

## 2 Answers

You have $$\mathbb{R}^4\setminus \{(0,0,0,w) | w\in \mathbb{R} \} = (\mathbb{R}^3\setminus \{(0,0,0 \}) \times \mathbb{R}$$, thus $$\mathbb{R}^4\setminus \{(0,0,0,w) | w\in \mathbb{R} \}$$ is homotopy equivalent to $$\mathbb{R}^3\setminus \{(0,0,0 \}$$. This space is homotopy equivalent to $$S^2$$ (in fact, $$S^2$$ is a strong deformation retract of $$\mathbb{R}^3\setminus \{(0,0,0 \}$$) which is known to be simply connected.

A way to visualize curves in $$\mathbb{R}^4$$ is to think about the curves as being in $$\mathbb{R}^3$$ through a projection, then keeping track of the $$w$$-coordinate through color along the curve. Two curves can pass through each other without intersecting in $$\mathbb{R}^4$$ if they have different colors when they intersect in $$\mathbb{R}^3$$.

In the projection onto the $$xyw$$ space, $$\mathbb{R}^4$$ minus a line is $$\mathbb{R}^3$$ minus a line. If you have a closed loop in the complement, you can unhook it from the line by changing the color of the loop so that it is a distinct color from the line.

In the projection onto the $$xyz$$ space, it's $$\mathbb{R}^3$$ minus the origin, in which case it's fairly obvious there's no obstruction to deforming any loop to a point!

Yet another way to see it is to deformation retract $$\mathbb{R}^4-\{(0,0,0,w):w\in\mathbb{R}\}$$ to the set of unit vectors in this space. This results in $$S^3$$ minus two points, which is simply connected.